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Question

1) $\mathop {\lim }\limits_{n \to +\infty }(\sqrt[3]{n^3+n^2}-\sqrt{n^2-1})$

2) $\mathop {\lim }\limits_{n \to +\infty }(\sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3})$

3) $\mathop {\lim }\limits_{n \to +\infty }\frac{n-\sqrt{n^2+n\sqrt{n}}}{\sqrt{n+1}-\sqrt{n+2}}$

4) $\mathop {\lim }\limits_{n \to +\infty }\frac{1+3+5+...+(2n-1)}{n^2+1}$

5) $\mathop {\lim }\limits_{n \to +\infty }\frac{n\sqrt{1+2+3+...+2n}}{3n^2+n-2}$

☼SunShine❤️ · Answer 1 · 1/13/2017 3:07:23 AM

5 )
Vì : $1+2+3+...+2n=2n^2 + n$
( Áp dụng CT tổng của CSC)
$\rightarrow lim_{n \to + \infty}(...)=lim_{n \to +\infty}\frac{n\sqrt{2n^2 +n}}{3n^2 +n -2}$
$=lim_{n \to + \infty}\frac{\sqrt{2+\frac{1}{n}}}{3+\frac{1}{n}-\frac{2}{n^2}}=\frac{\sqrt{2}}{3}$

Trả lời 13-01-17 03:07 AM

☼SunShine❤️
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:D à ừa chia rồi – meomeocutduoi 13-01-17 04:38 AM

xem lại đi =))) – ☼SunShine❤️ 13-01-17 04:18 AM

:D Sách giáo khoa trang 96 – meomeocutduoi 13-01-17 04:00 AM

chia 2 ? wtf ? – ☼SunShine❤️ 13-01-17 03:41 AM

:D quên chưa chia 2 chỗ tổng CSC kìa – meomeocutduoi 13-01-17 03:10 AM

☼SunShine❤️ · Answer 2 · 1/13/2017 2:49:59 AM

Bình chọn tăng 3 Bình chọn giảm

4 ) Vì : $1+3+5+...+(2n-1)=n^2$
( Có thể CM = pp quy nạp )
$\rightarrow lim_{n \to +\infty}(...)=lim_{n \to + \infty}\left ( \frac{n^2}{n^2+1} \right )=1$

Trả lời 13-01-17 02:49 AM

☼SunShine❤️
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:D cấp số cộng – meomeocutduoi 13-01-17 03:01 AM

☼SunShine❤️ · Answer 3 · 1/13/2017 2:47:11 AM

Bình chọn tăng 3 Bình chọn giảm

3) $lim_{n \to +\infty}\frac{n-\sqrt{n^2+n\sqrt{n}}}{\sqrt{n+1}-\sqrt{n+2}}$
$=lim_{n \to + \infty}(\sqrt{n^2+n\sqrt{n}}-n)(\sqrt{n+1}+\sqrt{n+2})$
$=lim_{n \to +\infty}\frac{n\sqrt{n}}{\sqrt{n^2+n\sqrt{n}}+n}\left ( \sqrt{n+2}+\sqrt{n+1} \right )$
$=lim_{n \to + \infty}(n.\frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt{n+\sqrt{n}}+\sqrt{n}})$
$=lim_{n \to +\infty}(n.\frac{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}}{\sqrt{1+\frac{1}{\sqrt{n}}}+1})$
$\Rightarrow lim_{n \to +\infty}(...)=+ \infty$

lịch sử

Sửa 13-01-17 02:47 AM

☼SunShine❤️
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Trả lời 13-01-17 02:42 AM

☼SunShine❤️
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☼SunShine❤️ · Answer 4 · 1/12/2017 10:40:41 PM

Bình chọn tăng 4 Bình chọn giảm

2 ) $lim_{n \to+ \infty}\left ( \sqrt{n^2+2n+3}+\sqrt[3]{n^2-n^3} \right )$
$=lim_{n \to+ \infty}\left ( \sqrt{n^2+2n+3}-n+n+\sqrt[3]{n^2-n^3} \right )$
$=lim_{n \to+ \infty}\left ( \frac{2n+3}{\sqrt{n^2+2n+3}+n} +\frac{1}{1-\sqrt[3]{\frac{1}{n}-1}+\sqrt[3]{1-\frac{2}{n}+\frac{1}{n^2}}} \right )$
$\Rightarrow lim_{n \to + \infty}\left ( ... \right )=1+\frac{1}{3}=\frac{4}{3}$

Trả lời 12-01-17 10:40 PM

☼SunShine❤️
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đúng thì Clck V dùm mk nhé ! ~~ cám ơn <3 – ☼SunShine❤️ 14-01-17 08:24 PM

rõ gì bạn ? – ☼SunShine❤️ 14-01-17 08:23 PM

bn ghi rõ đk ko – pokerface 14-01-17 08:08 AM

☼SunShine❤️ · Answer 5 · 1/12/2017 10:30:30 PM

Bình chọn tăng 4 Bình chọn giảm

1) $lim_{n \to +\infty} \left ( \sqrt[3]{n^3+n^2}-\sqrt{n^2-1}\right )=lim_{n \to +\infty}\left ( \sqrt[3]{n^3+n^2}-n + n-\sqrt{n^2-1} \right )$
$=lim_{n \to + \infty}\left ( \frac{n^2}{\sqrt[3]{n^6 +2n^5 + n^4}+\sqrt[3]{n^6+n^5}+n^2} +\frac{1}{n+\sqrt{n^2-1}}\right )$
$=lim_{n \to +\infty}\left ( \frac{1}{\sqrt[3]{1+\frac{2}{n}+\frac{1}{n^2}}+\sqrt[3]{1+\frac{1}{n}}+1}+\frac{\frac{1}{n}}{1+\sqrt{1-\frac{1}{n^2}}} \right )$
$\Rightarrow lim_{n \to +\infty}\left ( .... \right )=\frac{1}{3}$

lịch sử

Trả lời 12-01-17 10:30 PM

☼SunShine❤️
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